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OCRChemistryAH432 Acids,BasesandBuffers

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DefinitionsofacidbehaviorStrong:dissociatedfullytoproducehydrogenionsinaqueoussolutionWeak:partiallydissociatedtoproducehydrogenionsinaqueoussolutione.g. HClisastrongacid:HCl(g)!H+

(aq)+Cl-(aq)whendissolvedinwater soif1moleofHClisdissolvedin1dm3ofwater,the[H+]willbe1moldm-3 CH3COOHisaweakacid:CH3COOH(aq)⇌H+

(aq)+CH3COO-(aq)

if1moleofCH3COOHisdissolvedinwater,the[H+]willbemuchlessthan1moldm-3 andmostoftheCH3COOHmoleculeswillremainundissociated.ThesearebasedonArrhenius'definitionofanacid:producingH+ionsinaqueoussolution.Thedefinitionisusefulinmanycases,butnotall.e.g. HCl(g)+NH3(g)!NH4Cl(s) isalsoanexampleofanacidreactingwithabase. HeretheHClmoleculeisgivingH+totheammoniamolecule,butnoneofthisishappeninginsolution.BronstedandLowrygeneratedthemoregeneraldefinition: Acid=PROTONDONOR–candonateH+ Base=PROTONACCEPTOR–canacceptH+ConjugateacidsandbasesWhenanaciddonatesitsproton,theremainingspeciesisaprotonacceptor–i.e.abase–becausethereversereactioncantakeplace.Similarlywhenabaseacceptsaproton,thespeciesformedisaprotondonor–i.e.anacidsinceitcandonatetheprotoninareversereaction.Wecalltheseconjugateacid/basepairs.Wedefine:ThebaseformedwhenanacidlosesitsprotoniscalledtheCONJUGATEBASEofthatacid.e.g. Cl-istheconjugatebaseofHClTheacidformedwhenabasegainsaprotoniscalledtheCONJUGATEACIDofthatbase.e.g. NH4

+istheconjugateacidofNH3Wecanidentifyconjugateacid/basepairsinthereactionsofacidwithbases: HCl+ NH3 ! NH4

+ + Cl- acid baseconjugateacid conjugatebase (acid1) (base2)(acid2)(base1)


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Whenammoniagasisbubbledthroughwater: NH3(g) + H2O(l) ⇌ NH4

+(aq) + OH-

(aq) base acid conjugateacidconjugatebase (base1)(acid2)(acid1)(base2)Thisexplainswhyanammoniasolutionisalkaline.Whenhydrogenchlorideisbubbledthroughwater: HCl(g) + H2O(l) ! H3O+

(aq) + Cl-(aq) acid base conjugateacid conjugatebaseHerewehaveshownthehydroniumion,H3O+,whichishowhydrogenionsexistwhentheyareinwater.WeusuallyjustrepresentthisasH+

(aq),butweshouldbeawarethathydrogenions(protons)areneverfoundisolated,theyarealwaysbondedtosomethingbecausetheyare10,000timessmallerthananyotherchemicalspeciesandtheirsmallsizeandpositivechargegivesthemanintenseelectricfield,attractinganynearbylonepairtoformadativebond.

Practice:12) HBrisanacid.Whatisitsconjugatebase? Br-13) OH-isabase.Whatisitsconjugateacid? H2O14) Identifytheconjugateacid/basepairsinHNO3+OH-!NO3

-+H2O acidbaseconbconaRepresentingacidbehaviourThereactionsofacidsarebestshownusingionicequationsasthesefocusontheessentialreactantsandignorethespectatorions.ThereactionsofacidscanthenbeseentobereactionsoftheH+ion.

Steps:1)Writeafullbalancedequation2)Splitupanythingaqueousintoitscomponentions3)Cancelanyionswhichareunchangedonbothsidesoftheequation(spectatorions)

i)Withmetalse.g. Generalequation metal+acid!salt+hydrogen Balancedequation Mg(s)+2HCl(aq)!MgCl2(aq)+H2(g)

Step2 Mg(s)+2H+(aq)+2Cl-(aq)!Mg2+(aq)+2Cl-(aq)+H2(g)

Step3 Mg(s)+2H+(aq)!Mg2+(aq)+H2(g)

i.e.anyacidreactswithmagnesiumtoproducemagnesiumionsinsolutionandbubblesofhydrogen

ii)Withmetaloxidese.g. Generalequation metaloxide+acid!salt+water Balancedequation CuO(s)+H2SO4(aq)!CuSO4(aq)+H2O(l) Ionicequation CuO(s)+2H+

(aq)!Cu2+(aq)+H2O(l)i.e.anyacidreactswithcopperoxidetogivecopperionsinsolution


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iii)Withmetalhydroxidese.g. Generalequation metalhydroxide+acid!salt+water Balancedequation 2KOH(aq)+H2SO4(aq)!K2SO4(aq)+2H2O(l)

Step2 2K+(aq)+2OH-(aq)+2H+

(aq)+SO42-(aq)!2K+(aq)+SO4

2-(aq)+2H2O(l)

Ionicequation 2OH-(aq)+2H+

(aq)!2H2O(l)i.e.anyacidneutralisesanyhydroxide,producingwater

iv)Withmetalcarbonatese.g. Generalequation metalcarbonate+acid!salt+water+carbondioxide Balancedequation Na2CO3(aq)+2HNO3(aq)!2NaNO3(aq)+H2O(l)+CO2(g) Ionicequation: CO3

2-(aq)+2H+

(aq)!H2O(l)+CO2(g)i.e.anycarbonatereactswithanyacidtoproducecarbondioxideandwaterClassifyingacidsAcidsaredescribedintermsofthenumberofH+ionseachacidmoleculecanproduce: MonobasicacidsproduceoneH+ion,e.gHCl,HNO3,CH3COOH DibasicacidsproducetwoH+ions,e.g.H2SO4,HOOCCOOH TribasicacidsproducethreeH+ions,e.g.H3PO4,H3BO3,citricacidIonicProductofWaterWatermoleculescanreactwitheachotherwhentheycollide–averysmallproportionofthemexchangeaprotontoformions.Onewatermoleculeisactingasanacid,andtheotheractingasabase–waterisamphoteric. H2O+H2O⇌H3O++OH- althoughwecouldwriteH2O(l)⇌H+

(aq)+OH-(aq)

Wecanwriteanequilibriumconstantexpressionforthis: Kc=[H+

(aq)][OH-(aq)]units:mol2dm-6

(remember,whenapureliquidisinexcess,e.g.waterhere,wedon’tincludeitintheKcexpression).Becausethisisanexpressionspecifictowater,wecallthisequilibriumconstantKw,andrefertoitastheionicproductofwater.Likeanyequilibriumconstant,Kwdoesnotvarywithanythingexcepttemperature.Atroomtemperatureithasavalueof10-14mol2dm-6.Inpurewater,alltheH+andOH-areformedbywatermoleculessplitting,so[H+]=[OH-]ThismeansthatKw=[H+]2so[H+]=√Kw=10-7IfwerelatethistopurewaterhavingapHof7,wecanbegintoseethebasisofthepHscale,whichisameasureoftheconcentrationofH+ionsinasolution.ThepHscaleInventedbySorenSorensenofCarlsbergInstituteformeasuringacidityinbrewing.Definition: pH=-log10[H+]


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e.g. [H+] pH 10-2 2 10-4 4 10-7 7 10-11 11Note: 10-0=1 0WealsoneedtobeabletoturnapHintoaconcentrationofH+ions. [H+]=10-pH

UsingthepHscale:" thesmallerthepH,thehighertheconcentrationofH+

(aq)" pH0correspondstoa[H+]of1moldm-3" pH14correspondstoa[OH-]of1moldm-3(notethateveninaveryalkalinesolutionsuchas

thistherewillstillbea[H+]of10-14moldm-3)" adifferenceof1pHunitcorrespondstoaten-folddifferenceinacidity.pH2istentimes

moreacidicthanpH3.Practice:1) Inasolution,[H+

(aq)]=10-4moldm-3.Whatis[OH-(aq)]?10-10moldm-3

2) Anacidproducesan[H+(aq)]of0.1moldm-3.WhatisitspH?1.0

3) AnacidhasapHof3.0Whatisthe[H+(aq)]? 0.001moldm-3

FindingthepHofstrong(monobasic)acidsThesearefullydissociated,so[acid]=[H+]WesimplyusethepHdefinition: pH=-log10[H+]e.g.Ihavea0.02moldm-3solutionofHCl.WhatisitspH? [HCl]=0.02moldm-3HClisastrongacid,so[H+]=0.02moldm-3 pH=-log10(0.02) =1.70 (use2dpforpHvalues,typically)Note:Itispossibletohaveanacidwhichproducesmorethanoneprotonwhenitdissociates.Forexample H2SO4!2H+

(aq)+SO4-2(aq)

So[H+]=2x[H2SO4]!e.g. Ihavea0.02moldm-3solutionofH2SO4.WhatisitspH? [H2SO4]=0.02moldm-3andH2SO4isastrongacidso[H+]=2x0.02=0.04moldm-3 pH=-log10(0.04)=1.40Wealsoneedtobeabletocalculate[H+]forastrongacid:Weusetheformula[H+]=10-pHtocalculatethisdirectly.


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e.g.Whatisthe[H+]inasolutionofhydrochloricacidwithapHof0.7? [H+]=10-0.7=0.20moldm-3

Practice:4)WhatisthepHofasolutionofnitricacidofconcentration0.50moldm-3?0.305)Whatisthe[H+]inanitricacidsolutionwithapHof2.5?0.0032moldm-3FindingthepHofstrongalkalisHereweknowthealkaliisfullydissociatedso[OH-]=concentrationofthealkaliinsolution.WehavetousetheKw=[H+][OH-]expressiontoget[H+].(ValueofKwwillbegiven.)ThenwecancalculatepHasbefore.e.g.Ihavea0.2moldm-3solutionofNaOH.WhatisitspH? [NaOH]=0.2moldm-3andNaOHisastrongalkali,so[OH-]=0.2moldm-3 Kw=[H+][OH-]andKw=10-14mol2dm-6 so10-14=[H+]x0.2 so[H+]=10-14/0.2=5x10-14 pH=-log10(5x10-14)=13.3 Wecangettothissameresultaslightlydifferentroute: pH=14+log10[OH-] =14+log10(0.2)=14–0.7=13.3 Thisisactuallythesamecalculation,substitutingKwand[OH-]for[H+]inthepHequation andsimplifyingit–tryitifyoudon'tbelieve…Wealsoneedtobeabletocalculatethe[OH-]ifwearegiventhepH:WeusethepHtoget[H+],thenweuseKwtoget[OH-].e.g.Whatisthe[OH-]inasolutionofpotassiumhydroxidewithpH=11.3?(Kw=10-14) [H+]=10-pH=10-11.3=5.01x10-12 Kw=[H+][OH-]so10-14=5.01x10-12x[OH-] [OH-]=10-14/5.01x10-12=0.00200 Practice:6)CalculatethepHofa0.05moldm-3solutionofpotassiumhydroxide.(Kw=10-14) 10-14=[H+]x0.05=>[H+]=10-14/0.05=2x10-13 pH=-log10(2x10-13)=12.7 orpH=14+log10(0.05)=14–1.30=12.77)Whatisthe[OH-]inasodiumhydroxidesolutionwithpH9.0?(Kw=10-14) [H+]=10-9so10-14=10-9x[OH-]so[OH-]=1x10-5moldm-3


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pHofwaterWeexpectpurewatertobeneutral(whichitis)andtohaveapHof7.WethereforeerroneouslythinkthatneutralityisdefinedbypH=7,whichitisnot.Neutralmeans[H+]=[OH-].BecauseKwincreaseswithtemperature,athighertemperaturesmorewatermoleculesaresplitintoions(whichwe'dexpectonthebasisofmorefrequentenergeticcollisions).ThismeansthattheconcentrationsofbothH+andOH-arehigherinhotterwater.IftheconcentrationofH+ishigher,thepHislower(moreacidic)than7.Thewaterisstillneutral.CalculatingthepHofweakacidsRecallthataweakacidisonlypartiallydissociated,so[H+]islessthan[HA]wheretheacidisrepresentedbytheformulaHA.Inordertoworkoutwhatthe[H+]willbe,weneedtoknowhowmuchoftheacidhassplitupintoions.Theequilibriumconstanttellsusthis.Forweakacidswecallthistheaciddissociationconstant,Ka: HA(aq)⇌H+

(aq)+A-(aq)

Ka=[H+

(aq)]eqm[A-(aq)]eqmunits:moldm-3

[HA(aq)]eqmThelargerKais:-themoretheaciddissociatesintoions-thestrongertheweakacidis-themore[H+]inthesolution-thelower(moreacidic)thepHUsingKawecanworkout[H+]andthenuseittoworkoutthepH:TheH+ionshavebeenformedbyHAsplittingup,soforeveryH+ion,thereisanA-ion.Thismeans[H+]=[A-]atequilibriumsowecansubstituteintotheKaequation: Ka=[H+

(aq)]eqm[H+(aq)]eqm=[H+

(aq)]2eqmso[H+(aq)]2eqm=Kax[HA(aq)]eqm

[HA(aq)]eqm [HA(aq)]eqm and[H+

(aq)]eqm=√(Kax[HA(aq)]eqm)socancalculateH+,andhencethepH.LimitationsWeknowthevalueofKa,butwedon'tknow[HA]eqm.Solongastheacidisonlyaverysmallamountdissociated(i.e.solongasKaisverysmall),thenwecansaythat[HA]eqm≅[HA]undissociatedi.e.theconcentrationofHAatequilibriumisapproximatelythesameastheconcentrationofHAwestartwith. Thusforaweakacid:[H+]≅√(Kax[HA])Thisapproximationwillholdsolongastheweakacidisn’tverystrong–ifitdissociatesmore,then[HA]undissociatedwillbelargerthan[HA]eqmandso[H+]eqmwillbetoolargeandpHtoolow.


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Theotherapproximationwehavemadeisthatallthe[H+]camefromthedissociationofHA.Weused[H+]eqm=[A-]eqmbutweknowthatH2OalsodissociatesaverysmallamountintoH+andOH-so[H+]eqmwillbealittlebitlargerthan[A-]eqm.Theapproximationisfinesolongastheweakacidisn’textremelyweak(i.e.Kaisn’textremelysmall).WhenthereisonlyatinyamountofH+fromdissociationoftheacid,theamountofH+fromdissociationofwaterbecomessignificant.Whentheacidisveryweak,thecalculatedvalueof[H+]eqmcomesouttoosmall,andhencepHtoohigh.e.g.CalculatethepHofa0.1moldm-3solutionofethanoicacid. (Kaforethanoicacid=1.7x10-5moldm-3) [HA]=0.1moldm-3 [H+]≈√(Kax[HA]) =√(1.7x10-5x0.1)=0.0013 pH=-log10(0.0013) =2.88Practice:calculatethepHof0.0200moldm-3ofethanoicacid.Ka=1.74x10-5moldm-3 3.23calculatethepHof0.0100moldm-3butanoicacid.Ka=1.51x10-5moldm-3 3.41calculatethepHof0.500moldm-3chloroethanoicacid.Ka=1.38x10-3moldm-3 1.58FindingKaforaweakacidWemightalsobeexpectedtoworkoutKaforaweakacid,givenitspHandtheconcentrationoftheundissociatedacid.Wecanusetheassumptionsweusedbefore,that[H+]eqm=[A-]eqmandthat[HA]eqm=[HA]undissociated,solongasweknowwhenthoseapproximationsbreakdown.e.g.A1.0moldm-3solutionofethanoicacidhasapHof2.38.WhatisthevalueofKaforethanoicacid? [H+]=10-pH=10-2.38=0.00417 Ka=[H+]eqm[A-]eqm/[HA]eqmbut[A-]=[H+]and[HA]eqm=[HA]=1.0moldm-3 soKa=[H+]2/1.00=(0.00427)2=1.74x10-5moldm-3 NotealsothatKa(likeanyequilibriumconstant)changeswithtemperature,sowereallyoughttostatethetemperatureatwhichthepHvaluesweremeasured,andquotethevalueofKaasvalidatthistemperature.pKaThenumericvaluesofKaforweakacidaresomewhatcumbersome,sosometimespKaisquotedinsteadforweakacids.ItrelatestoKainthesamewayaspHdoesto[H+]. pKa=-log10Kaor,rearrangingKa=10-pKaThesmallerthevalueofpKais,thelargerthevalueofKaandthereforethemoreionizedtheweakacidis–andsothestrongertheweakacidis.


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e.g. acid Ka pKa chloroethanoic 1.3x10-3 2.88 benzoic 6.3x10-5 4.20 ethanoic 1.7x10-5 4.77 hydrocyanic 4.9x10-10 9.31e.g. WhatisthepHofa0.1moldm-3chloroethanoicacidsolution? pKa=2.88soKa=10-2.88=1.3x10-3 [H+]≈√(Kax[HA])so[H+]=√(1.3x10-3x0.1)=0.0114 pH=-log10[H+]=-log(0.0114)=1.94Practice:11)WhatisthepHofa0.01moldm-3solutionofbenzoicacid,forwhichpKa=4.20? [H+]=0.00079sopH=3.1MeasuringpHTheonlyreliablewayofmeasuringpHdirectlyistouseapHprobe.Universalindicatoronlyprovidesanapproximation.ApHprobeneedstobeclean,andcalibrated.ThisisdoneusingsolutionsofknownpHwhicharecapableofresistingtheadditionofsmallamountsofacidoralkaliwithonlyminimalchangestotheirpH.Wecallsuchsolutionsbuffers.BufferSolutionsAbuffersolutionisdefinedas:"asolutionwhichminimizes(orresists)changesinpHwhenasmallamountofacidoralkaliisadded."Thekeyideashereare:i)itminimizesorresiststhepHchange–not"maintainsaconstantpH,orcancelsoutthechange".ThepHdoeschange,butonlybyasmallamount–notasmuchasitwouldifthesmallamountofacid/alkaliwereaddedtoasolutionwhichwasn'tabuffer.ii)asmallamountofacidoralkalicanbeaddedbutthebufferingeffectdoesnotcopewiththeadditionoflargeamountsofacidoralkali.Buffersareveryimportantsystems–includinginbiochemistry.-bloodisabuffersolution.-manyskincareproductsarebufferedatpH5.5(thepHofnormalhealthyskin)-citricacidandsodiumcitrateareoftenusedinfoods,listedas"acidityregulators"-detergentsandshampoosareoftenbufferedtostayslightlyalkalineBuffersaremadebydissolvingaweakacidalongwithitsconjugatebase.Asolublesaltoftheacidprovidesthesupplyofconjugatebaseions.


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e.g.ethanoicacidisaweakacid. CH3COOH ⇌ CH3COO-+H+ whatistheconjugatebase?Agoodsourceofethanoateionswouldbesodiumethanoate.Ionicsaltslikethisdissociatefullyinaqueoussolution.Thebuffersolutionthereforewouldbemadebydissolvingethanoicacidandasolublesaltofethanoicacid(e.g,sodiumethanoate)inthesolution.Anotherwayofmakingthesamebufferwouldbetostartwithanethanoicacidsolution,andtopartiallyneutralizethesolutionusingsodiumhydroxide.Thiswouldformsomesodiumethanoateinthesolution,andsolongassomeoftheethanoicacidremainedinexcess,abufferwouldbeformed.Practical:Taketwobeakersofdistilledwater(100cm3)withpHprobesineach;addaspatulaofCH3COONatooneandexplainthemorealkalinepH.Addethanoicacid(100cm30.1M)tothesodiumethanoatetomakeabuffer.ExplainmoreacidicpH.AddafewdropsofHCltobothbeakersandcomparepHchanges.CountthenumberofdropsofbenchNaOHittakestomakethedistilledwaterpH11,andseewhatthesamenumberofdropsofNaOHdotothebuffer.Washandputawayprobes.Addfull-rangeUItoeachbeakerandcontinuetocompareeffectonindicatorcolouronaddingsmallamountsofacidandalkalitoeachbeaker.Finally'break'thebufferbyaddingmoreacid/alkaliandseeingtheindicatorinthebufferchange.IftheweakacidiswrittenasHA,thentheconjugatebaseisA-.Howeveritismade,bufferscontaini)ahighconcentrationofHAbecausemostoftheweakacidmoleculesremainundissociated HA(aq)H+

(aq)+A-(aq) Theequilibriumliesfartotheleft.

ii)ahighconcentrationofA-(theconjugatebase) becauseaswellthefewA-fromthedissociatedHA,wealsoaddA-ionsintheformofe.g.the sodiumsalt.iii)alowconcentrationof[H+] fromthefewdissociatedHAmolecules,andbecausetheaddedA-ionsshiftthepositionof the HA⇌H++A-equilibriumfurthertotheleft.HowdoesabufferresistchangesinpH?Thebuffersolutionworksbecausethepositionofequilibriumcanmoveleftorright: HA(aq)⇌ H+

(aq)+A-(aq)

Ifacidisaddedtothebuffer,thisisdealtwithbytheconjugatebaseA-ions.

" theconcentrationofH+(aq)increases

" thepositionofequilibriumshiftstothelefttouseuptheaddedH+ionstominimizetheeffectofthechange(LeChatelier)

" thisisdonebytheconjugatebase,A-(aq)reactingwiththeH+

(aq)toformmoreHA(aq)


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Ifanalkaliisaddedtothebuffer,thisisdealtwithbytheweakacid,HA." theconcentrationofOH-

(aq)isincreased" thesmallconcentrationofH+reactswiththeOH-ions,neutralizingthem: H+

(aq)+OH-(aq)!H2O(l)

" becauseofthedecreaseinH+(aq)concentration,thepositionofequilibriummovestothe

right,dissociatingmoreHAandreplacing(mostof)theH+ionsinthesolution(minimizingtheeffectofthechange–LeChatelier)

Thereisalimittohowmuchbufferingasolutioncando:

" ItcannotbufferagainstOH-ionsanymoreoncetheHAhasallbeendissociated" ItcannotbufferagainstH+ionsanymoreoncealltheA-hasbeenusedup

CalculatingthepHofabufferThepHofabuffercanbecalculatedusingtheequilibriumconstantexpressionfortheweakacid: Ka=[H+]eqm[A-]eqm [HA]eqm andasbeforewecanapproximate[HA]eqm≅[HA]undissociated

Similarlybecausewehaveaddedalotof[A-]ions,theadditional[A-]ionsgeneratedbythedissociationofHAwillbeinsignificant,so[A-]eqm≅[A-]initial

Thereforerearranging,[H+]≅Kax[HA] [A-]Where[HA]istheconcentrationoftheacidfromwhichyoumadethebufferand[A-]istheconcentrationofthesaltofthatacid(conjugatebaseions)youadded.Onceyouknow[H+]thepHissimplycalculatedasnormal:pH=-log10[H+]Practicingwiththismethod:1)Abufferismadetocontain0.2moldm-3ethanoicacidand0.1moldm-3sodiumethanoateinthesolution.WhatpHwillthebufferhave?(Kaforethanoicacid=1.7x10-5moldm-3) [H+]=1.7x10-5x0.2=3.4x10-5 0.1 pH=-log10(3.4x10-5)=4.472)Abufferismadeupbyadding200cm3of0.1moldm-3benzoicacidsolutionto500cm3of0.1moldm-3sodiumbenzoatesolution.WhatpHwillthebufferhave?(Kaofbenzoicacid=6.3x10-5moldm-3) molesofbenzoicacid=0.1x(200/1000)=0.02 molesofsodiumbenzoate=0.1x(500/1000)=0.05


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totalvolumeofsolution=200cm3+500cm3=700cm3 [benzoicacid]=0.02/0.7=0.02857moldm-3 [sodiumbenzoate]=0.05/0.7=0.0714moldm-3 [H+]=6.3x10-5x0.02857 =2.52x10-5 0.0714 pH=-log10(2.43x10-5)=4.60(4.5984…)socheckroundingerrors 3)250cm3of0.1Msodiumhydroxidesolutionareaddedto750cm3ofethanoicacidsolutionof0.1Mconcentration.WhatwillbethepHoftheresultingbuffer?(pKaforethanoicacid=4.8) molesofNaOH=0.25x0.1=0.025 molesofCH3COOH=0.75x0.1=0.075 Intheneutralization,NaOH+CH3COOH⇌CH3COO-Na++H2O So0.025molesofNaOHareusedtoneutralize0.025molesofCH3COOH,creating0.025 molesofsodiumethanoate,andleaving0.050molesofethanoicacid Thetotalvolumeofthesolutionisnow250cm3+750cm3=1000cm3(1dm3) So [CH3COOH]=0.050x1.0=0.05moldm-3 [CH3COO-]=0.025x1.0=0.025moldm-3 Ka=10-pKa=10-4.8=1.58x10-5

[H+]=1.58x10-5x0.05 =3.16x10-5 0.025 pH=-log10(3.16x10-5)=4.5Practice:CalculatethepHat25°Cofabuffersolutionmadeupfrom250cm3of0.10moldm-3ethanoicacidand250cm3of0.20moldm-3sodiumethanoate.(Kaforethanoicacid=1.7x10-5moldm-3at25°C) [H+]=1.7x10-5x0.05=8.5x10-6 0.1 pH=-log10(8.5x10-6)=5.074.42gofsodiummethanoate,HCOO-Na+,isdissolvedin1.0dm3ofmethanoicacidofconcentration0.15moldm-3at25°C.WhatwillthepHoftheresultingbufferbe?(Ka=1.6x10-4moldm-3formethanoicacidat25°C) HCOONa=1+12+32+23=RFM68 moles=4.42/68=0.065 so[A-]=0.065moldm-3 [HA]=0.15moldm-3 So[H+]=1.6x10-4x0.15 =3.69x10-4 0.065 pH=-log10(3.69x10-4)=3.43


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DesigningabufferforagivenpHWemayneedtousethismethod"inreverse"togofromaspecifiedpHtoa"recipe"forHAandA-concentrationstousetomakeabufferofaspecificpH.Example:SuggestconcentrationsofethanoicacidandsodiumethanoatewhichwouldresultinabuffersolutionwithapHof4.07at25°C.(Kaforethanoicacid=1.7x10-5moldm-3atthistemperature). [H+]=10-pH so[H+]=10-4.07=8.51x10-5moldm-3

Ka=[H+][A-] WeknowKaandH+,sowecanrearrangethis: [HA] [HA]=[H+]x[A-] inthiscase[HA]=8.51x10-5x[A-] Ka 1.7x10-5 [HA]=5.00x[A-]Thereforeweneedaconcentrationoftheacidwhichisfivetimestheconcentrationofthesaltoftheweakacid.Wecanmakethebufferusinganyconcentrationswelike,solongastheyareinthisratio.Perhaps0.5MHAand0.1MA-wouldbechosen.BufferinginbloodHealthybloodplasmarequiresapHbetween7.35and7.45.IfthepHdropsbelow7.2orabove7.6thebodyisinserioustrouble.Abufferingsystemisthereforeneededwithinthebodytodealwithsmallamountsofacidoralkalienteringthebloodstream.ThepHofbloodiscontrolledbyamixtureofbuffers.Themostimportantoftheseisthecarbonicacid–hydrogencarbonateionbuffer.Carbonicacidisformedinthebloodwhencarbondioxidedissolvesinwater.Itisaweakacid: H2CO3(aq)⇌H+

(aq)+HCO3-(aq)

Thehydrogencarbonateion,HCO3

-isitsconjugatebase,andispresentinbloodinsignificantconcentrations.Anyincreasein[H+

(aq)]inthebloodisdealtwithbytheHCO3-ions(theconjugatebase)

• theaddedH+ionsreactwiththehydrogencarbonateionsinthebloodtoformH2CO3• becausetheequilibriumshiftstothelefttoremove(mostof)theH+whichwasadded(Le

Chatelier)Anyincreasein[OH-(aq)]inthebloodisdealtwithbytheH2CO3(theweakacid)• theaddedOH-ionsreactwiththefewH+ionsinthebloodtoformH2O• thepositionofequilibriummovesright,dissociatingH2CO3toreplace(mostof)theH+ionsinthe

blood.


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Acommonquestionistocalculatetheconcentrationsofcarbonicacidandhydrogencarbonateionsinblood.TheratioofHCO3

-:H2CO3canbecalculatedusingtheHendersonequationandtheKavalueforcarbonicacid(4.17x10-7moldm-3),andthepHoftheblood.Forhealthyblood,pH=7.4 pH=pKa–log([HA]/[A-]) pKa=-log(4.17x10-7)=6.38 7.40=6.38–log([HA]/[A-]) solog([HA]/[A-])=-1.02 and[HA]/[A-]=10-1.02=0.095 Theratio[HA]:[A-]=0.095:1or1:10.5(dividingbysmallest)Theconcentrationofhydrogencarbonateionsinhealthybloodismorethantentimestheconcentrationofcarbonicacid,inordertobufferatthecorrectpH.ReactionsofweakacidsTestyourunderstanding:i)Inatitration,20cm3of0.1MNaOHwererequiredtoneutralize25cm3ofahydrochloricacidsolution.Anethanoicacidsolution(Ka=1.7x10-5moldm-3)ofthesameconcentrationwasprepared.Whatvolumeof0.1MNaOHwouldberequiredtoneutralize25cm3ofit?Ans:20cm3.ii)A1gchunkofcalciumcarbonatewasaddedto100cm3of0.5Mhydrochloricacid.

• whatisthe[H+]intheacid? 0.5moldm-3• whichreactantisinexcess? RFMCaCO3=40+12+48=100=>0.01moles

MolesofHCl=0.1x0.5=0.05#excess (0.02stochiometric)

• whatwouldyouobserve? effervescence calciumcarbonatewouldalldissolveAsimilar1gchunkofcalciumcarbonatewasaddedto100cm3of0.5Methanoicacid.(Kaforethanoicacid=1.7x10-5moldm-3)

• whatisthe[H+]intheacid? [H+] =√(Kax[HA]) =√(1.7x10-5x0.5) =0.003mol

• whichreactantisinexcess? stilltheacid–thereare0.05Mofethanoicacid eventhoughthereareonly0.0003molesofH+

• whatwouldyouobserve? bubblesproducedMORESLOWLY allthecalciumcarbonateSTILLdissolves


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Whenaweakacidreacts,itreactsmoreslowlythanastrongacidofthesameconcentration.Thisisbecause[H+]ismuchlower,sotherearefewersuccessfulcollisionsinvolvingthe[H+]ion.Everymoleoftheacidwillreact,however.Asthe[H+]ionsareusedup,more[HA]splitsuptoproduceions,attemptingtorestoretheequilibrium(LeChatelier'sprinciple).EventuallyalltheHAmoleculeswillhavedissociatedandalltheH+ionsreacted.Thismeansthatthesamevolumeofalkalineutralizesastrongandaweakacidofthesameconcentration(sotitrationcan’ttellaweakfromastrongacidofthesameconcentration).UsingenthalpychangeofneutralisationWecanmeasuretheenthalpychangewhenanacidisneutralizedusinganalkalibymeasuringatemperaturechangeandthenusingQ=mcΔTandΔH=-Q/molesDefinition:Thestandardenthalpychangeofneutralizationistheenthalpychangethatoccurswhen1moleofwaterisproducedduringthereactionofanacidwithanalkaliunderstandardconditions.e.g. HCl+KOH!KCl+H2O ΔHө

neut=-57.9kJmol-1

HNO3+NaOH!NaNO3+H2O ΔHөneut=-57.6kJmol-1

Infactwefindthatforanystrongacidandanystrongalkali,thevalueisthesame.Thisisnotunexpected–wecanreducealloftheseequationstoionicequationsbyremovingthespectatorions.Whenwedowefindtheyareallthesamereactiontakingplace: H+

(aq)+OH-(aq)!H2O(l)

Evenifweusesulphuricacid,whichproducestwohydrogenionsinsolutionratherthanone,theresultisthesamebecausethedefinitionisnotbasedonmolesofacidoralkalibuton1moleofwaterbeingformed: ½H2SO4+NaOH!½Na2SO4+H2O ΔHө

neut=-57.9kJmol-1Ifwemeasurethestandardenthalpychangeofneutralizationwhenaweakacidsuchasethanoicacidreactswithastrongalkali,e.g.sodiumhydroxide,wefindthatΔHө

neutissmaller(-57.1kJmol-1)Sincethereactionwhichistakingplaceisthesame,thedifferencemustlieinethanoicacidbeingweakratherthanstrong.WhatweareseeingistheenergyneededtodissociatetheHAmoleculestoformH+ions.CH3COOH⇌CH3COO-+H+ΔH=+ve(onlybondbreaking,nomaking)AstheH+isneutralizedbythealkali,thepositionofequilibriummovestotheright,andenergythatwouldotherwisehavebeengivenoutasheathastobeusedtobreakH-Abonds.


p.15

Similarlywithaweakbase(e.g.ammonia),energyneedstobesuppliedtodissociatethebaseandproduceOH-ions,sotheenthalpychangeonneutralizationislower.Experimentsofthistypegivesomemeasureofhowstrongorweakacidsorbasesare,buttheyarenotparticularlyaccurate.UsingindicatorsTheuseofindicatorstocomparestrengthsofacidsmayappearcounter-intuitive.Surelyanindicatorwouldonlytellyouwhethersomethingisacidicornot?Toseehowthiscanbedoneweneedtounderstandinmoredepthwhatindicatorsactuallyare,andhowtheywork." Indicatorshavetwocolours,andchangebetweenthesetwocoloursdependingonthepH." ThepHatwhichanindicatorchangescolourisNOT(usually)pH7.0" TheychangecolouroverapHrange(typicallyaround2pHunits).Overthisrange,theintensityof

onecolourisfadingandtheintensityoftheotherisincreasing,sothecolourischanginggradually.Inthemiddleoftherange,thecolourisanequalmixtureofthetwo

Indicatorsareweakacids,andthereforepartiallydissociatedinaqueoussolution.Theundissociatedacidisonecolourinsolution,andtheconjugatebaseistheothercolour.e.g. HIn ⇌H++In- red yellow e.g.methylorangeIfweaddH+(acid)tomethylorange,thepositionofequilibriummovestothelefttoreducetheconcentrationofH+andthereismoreoftheredformandlessoftheyellowformi.e.theindicatorturnsredIfweaddOH-(alkali)tomethylorange,thisreactswiththeH+formingwaterandreducingtheH+concentration.ThepositionofequilibriummovesrighttocreatemoreH+,usinguptheredformoftheindicatorandcreatingmoreoftheyellowform.MethylorangehasapublishedpHrangeof3.2–4.4.ThismeansthatatpHlowerthan3.2itisred.FrompH3.2topH4.4itbecomeslessredandmoreyellow.ItwillbeorangeatpH3.8(themid-rangepoint).


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Bychoosingdifferentindicators,itispossibleto"pindown"thepHofaweakacidsolutiontoarangeofpHvalues,andthenusingthissuggestedpHtocalculate[H+]and,iftheconcentrationoftheacidisknown,anestimateforKa.Workedexample:BromocresolgreenhasapHrangeof3.8(yellow)to5.4(blue).MethylorangehasapHrangeof3.2(red)to4.4(yellow).i)AsolutionofweakacidHAwith0.1moldm-3concentrationistestedwithbothindicators.Withbothindicatorsthecolourisseentobeyellow.SuggestthepHoftheacidsolution.WithbromocresolgreentheyellowindicationtellsusthatthepHislessthanthemidpointoftherange.Themidpointis4.6(atwhichthecolourwillbegreen)sowecansaythatthepHofouracidis4.5orless.WiththemethylorangetheyellowindicationtellsusthatthepHishigherthanthemidpointoftherange.Themidpointisat3.8sowecansaythatthepHofouracidmustbe3.9ormore.WethereforesuggestthepHofoursolutionisavaluebetween3.9and4.5.Perhapswechoose4.2ii)UsingyoursuggestedpH,findKafortheacidHA. [H+]=10-pH=10-4.2=6.3x10-5moldm-3 Ka=[H+][A-] andforaweakacidsolution[H+]=[A-] [HA] and[HA]isapprox.thesameatequilibriumasintheundissociatedacid Ka=6.3x10-5x6.3x10-5 =3.9x10-8moldm-3 0.1iii)Anotheracid,HZ,havingthesameconcentrationisalsotested.Itturnsbromocresolgreenyellow,andmethylorangered.SuggestapHforthisacid.Ans:Withbromocresolgreen,theyellowcolourtellsusouracidhasapHof4.5orless(midpoint=4.6).WithmethylorangetheredcolourtellsusouracidhasapHof3.7orless(midpoint3.8)CombiningthisdataweknowthepHis3.7orless,buthowdoweputalowerlimitonthepH?Rememberthisisaweakacid.IfitwereastrongaciditspHwouldbe1.0,sothepHdefinitelycan'tbelowerthan1.0ifour0.1moldm-3acidisonlypartiallydissociatedinthesolution.OnestrategyistopickthepHthatcorrespondstotheendofthepHrangefortheindicatorwhosemidpointweusedtosettheotherlimit.InthiscasethepHislowerthanthemidpointformethylorange,sowe'llchoosetheacidicendofthepHrangeformethylorange;apHvalueof3.2asoursuggestedvalue.


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Acid-BaseTitrationsWhenweaddacidtoanalkali,orviceversa,(e.g.duringatitration)thepHdoesnotchangeinaconstantfashion.ThewayinwhichthepHchangesasacid/alkaliisaddedisshownusingapHcurve.Weneedtobeabletoexplain/interprettheshapesofthesecurves,andtosketchthem.Theshapeisdifferentdependingonwhethertheacidandthealkaliarestrongorweak.SketchingpHcurves(notingthekeymarkingpoints)1)Strongacid,towhichastrongalkaliisadded

i)agradualincreaseinpHfromthatoftheoriginalacidii)asuddensharp(vertical)changeinpHiii)centreoftheverticalpartcorrespondstopH7iv)agradualincreaseinpHtothatofthe(diluted)alkaliv)Theverticalsectioncorrespondstothatadditionofequalmolesofacid(H+)andalkali(OH-)somakesurethatthevolumesofacidandalkaliaddedreflectthise.g.equalvolumesifequalconcentrations.

Definition:Theequivalencepointisdefinedasthepointinatitrationatwhichthevolumeofacidhasexactlyreactedwiththeaddedalkali.ItthereforecorrespondstothecentreofthesteepestsectionofthepHcurve.2)Strongacid,towhichaweakalkalihasbeenadded

i)Theacidicpartofthecurveisthesameasbeforeii)TheverticalpartextendsoverfewerpHunitsiii)Thecentreoftheverticalsection(equivalencepoint)correspondstoanacidicpHiv)pHincreasesmoregraduallythroughoutalkalinepartv)finishingatthepHoftheweakalkalivi)makesureverticalsectionisatequimolarvolumes!

3)Weakacidtowhichastrongalkalihasbeenaddedi)TheacidicpartstartsatthepHoftheweakacidii)pHincreasessteadilyoveracidicsectioniii)theverticalsectionextendsoverfewerpHunitsiv)Thecentreoftheverticalsection(equivalencepoint)correspondstoanalkalinepHv)pHonlychangesverygraduallywithfurtheradditionofalkali,finishingatalkali'spHvi)makesureverticalsectionisatequimolarvolumes


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4)Weakacidtowhichweakalkalihasbeenadded

i)TheacidicpartstartsatthepHoftheweakacidii)pHincreasessteadilyiii)thereisanindistinctequivalencepoint(noverticalsection)correspondingtoapproximatelypH7iv)pHcontinuestorisesteadilyoveralkalinepartv)finishingatthepHoftheweakalkalivi)"inflexion"atequimolarvolumes

InterpretingapHcurve:1)Lookforthenatureoftheequivalencepoint. -isitindistinct(weakacidweakalkali)? -isitatpH7anddistinct(strongacidstrongalkali) -isitatacidicpH(strongacidweakalkali) -isitatalkalinepH(weakacidstrongalkali)2)Lookforthevolumecorrespondingtotheequivalencepointinordertoworkouthowmanycm3ofalkalihavebeenaddedtotheacid3)Rememberwecouldstartwiththealkaliandaddtheacid,inwhichcasethecurveswouldbemirrorimages.ChoosingindicatorsfortitrationsThechoiceofasuitableindicatorforagiventitrationisnotrandom.Definition:Theendpointofatitrationiswhentheindicatorismidwaybetweenitstwocolours,i.e.hasequalconcentrationsofHInandIn-.ThisisthemidpointofthepHrangefortheindicator.Inchoosinganindicatorforatitration,weneedtoknowwheretheequivalencepointwillbe.Wewanttheindicatortochangecolourwhenwereachtheequivalencepoint,sowewanttheendpointandtheequivalencepointtoco-incide.e.g.ifwehaveanequivalencepointatacidicpH,wewillneedtochooseanindicatorwithapHrangewhichisacidictoo.Inordertogetasharpendpoint–i.e.fortheindicatorcolourtochangewiththeadditionofonlyadropofacidoralkali,thenweneedtoensurethatthepHrangefortheindicatorcorrespondstotheverticalportionofthepHcurveforthattitration.Ifnot,theindicatorwillchangecolourgraduallyandwewon’tbeabletodetermineaccuratelyhowmuchacid/alkalitoadd.Wecanillustratethisbyoverlayingthecolourchangesoftwocommonindicators(phenolphthaleinandmethylorange)overthepHcurvesforstrongandweakacidandalkalititrations


p.19

1)Ifwearetitratingastrongacidagainstastrongbase:ThecolourchangesofbothindicatorsoccurswithintheverticalsectionofthepHcurveEitherindicatorwouldbeasuitablechoice2)Ifwearetitratingastrongacidagainstaweakbase:ThecolourchangeformethylorangelieswithintheverticalportionofthepHcurvei.e.theendpointiswell-matchedwiththeacidicequivalencepoint.ItwouldbeasuitablechoiceThecolourchangeforphenolphthaleinoccursgraduallyduringtheadditionofalkalilongaftertheequivalencepointhasbeenreached,sophenolphthaleinwouldbeunsuitable.3)IfwearetitratingaweakacidagainstastrongbaseThecolourchangeformethylorangelieswithinthegraduallychangingportionofthepHcurvei.e.thecolourchangesgraduallyasacidisadded,andhaschangedlongbeforetheequivalencepointisreached.Methylorangecannotbeused.ThecolourchangeforphenolphthaleinoccursduringtheverticalsectionofthepHcurveandsotheendpointiswellmatchedwiththepHoftheequivalencepoint.Phenolphthaleincanbeused.4)WeakacidandweakalkaliBecausethereisnoverticalsectionofthepHcurveextendingoverapprox.2pHunits,noindicatorwillbesuitableasthecolourofanyindicatorwouldbechanginggradually.

UniversalIndicatorUniversalindicatorisaMIXTUREofseveraldifferentindicators,eachchosensoitscolourchangeoveradifferentpHrange,sothecolouroftheuniversalindicatorchangesseveraltimesoverawiderangeofpHvalues.UniversalindicatoristhereforeNEVERsuitablefortitrations!

definitions of acid behavior - wordpress.com...representing acid behaviour the reactions of acids...

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